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Prévia do material em texto

2-1
CHAPTER 2
The correspondence between the problem set in this fifth edition versus the
problem set in the 4'th edition text. Problems that are new are marked new and
those that are only slightly altered are marked as modified (mod).
New Old New Old New Old
1 4 mod 21 13 41E 33E mod
2 new 22 14 42E 34E mod
3 new 23 15 43E 35E
4 7 mod 24 17 44E 36E
5 2 mod 25 18 45E 37E
6 new 26 new 46E 38E
7 new 27 19 47E 39E
8 new 28 20 48E 40E
9 5 mod 29 21 49E 41E
10 6 30 22
11 8 mod 31 23
12 new 32 24
13 9 mod 33 new
14 10 mod 34 25 mod
15 11 35 26 mod
16 new 36 27 mod
17 new 37 28
18 16 mod 38 29
19 new 39E 31E mod
20 12 40E 32E
2-2
2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665
m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational
field ? How much mass can a force of 1 N support ?
Solution:
ma = 0 = ∑ F = F - mg
F = mg = 2 × 9.80665 = 19.613 N
 F = mg => m = F/g = 1 / 9.80665 = 0.102 kg
2.2 A model car rolls down an incline with a slope so the gravitational “pull” in the
direction of motion is one third of the standard gravitational force (see Problem
2.1). If the car has a mass of 0.45 kg. Find the acceleration.
Solution:
ma = ∑ F = mg / 3
a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2
2.3 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5
seconds. If the total car and driver mass is 1075 kg. Find the necessary force.
Solution:
Acceleration is the time rate of change of velocity.
ma = ∑ F ; a = dV / dt = (60 × 1000) / (3600 × 5) = 3.33 m/s2
Fnet = ma = 1075 × 3.333 = 3583 N
2.4 A washing machine has 2 kg of clothes spinning at a rate that generates an
acceleration of 24 m/s2. What is the force needed to hold the clothes?
Solution:
F = ma = 2 kg × 24 m/s2 = 48 N
2.5 A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a
speed of 75 km/h. What are the force and total time required?
Solution:
a = dV / dt => ∆t = dV/a = [ ( 75 − 20 ) / 4 ] × ( 1000 / 3600 )
∆t = 3.82 sec ; F = ma = 1200 × 4 = 4800 N
2-3
2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What
force is needed and what is the final velocity?
Solution:
Constant acceleration can be integrated to get velocity.
a = dV / dt => ∫ dV = ∫ a dt => ∆V = a ∆t = 3 × 10 = 30 m/s
V = 30 m/s ; F = ma = 950 × 3 = 2850 N
2.7 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2
kN now accelerates this system. What is the acceleration?
Solution:
ma = ∑ F ⇒ a = ∑ F / m
m = msteel + mpropane = 15 + (1.75 × 44.094) = 92.165 kg
a = 2000 / 92.165 = 21.7 m/s2
2.8 A rope hangs over a pulley with the two equally long ends down. On one end you
attach a mass of 5 kg and on the other end you attach 10 kg. Assuming standard
gravitation and no friction in the pulley what is the acceleration of the 10 kg mass
when released?
Solution:
Do the equation of motion for the mass m2 along the
downwards direction, in that case the mass m1 moves
up (i.e. has -a for the acceleration)
 m2 a = m2 g − m1 g − m1a
 (m1 + m2 ) a = (m2 − m1 )g
This is net force in motion direction
 a = (10 − 5) g / (10 + 5) = g / 3 = 3.27 m/s2
g
1
2
2.9 A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration
of 2 m/s2 relative to the ground at a location where the local gravitational
acceleration is 9.5 m/s2. Find the required force.
Solution:
F = ma = Fup − mg
Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N
2-4
2.10 On the moon the gravitational acceleration is approximately one-sixth that on the
surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface
on the moon. What is the expected reading? If this mass is weighed with a spring
scale that reads correctly for standard gravity on earth (see Problem 2.1), what is
the reading?
 Solution:
 Moon gravitation is: g = gearth/6
m m 
m m 
 Beam Balance Reading is 5 kg
 This is mass comparison
 Spring Balance Reading is in kg units
 length ∝ F ∝ g
 Reading will be 
5
6
 kg
 This is force comparison
2.11 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500-
L tank. Find the specific volume on both a mass and mole basis (v and v ).
Solution:
v = V/m = 0.5/1 = 0.5 m3/kg
v = V/n = 
V
m/M
 = Mv = 32 × 0.5 = 16 m3/kmol
2.12 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest
of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall
(average) specific volume.
Solution:
mair = ρ V = ρair ( Vtot − mgranite / ρ )
 = 1.15 [ 5 - (900 / 2400) ] = 1.15 × 4.625 = 5.32 kg
v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg
2.13 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800
kg/m3. If the system is decelerated with 6 m/s2 what is the needed force?
Solution:
m = mtank + mgasoline = 15 + 0.3 × 800 = 255 kg
F = ma = 255 × 6 = 1530 N
2-5
2.14 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid
inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity,
find the piston mass that will create a pressure inside of 1500 kPa.
Solution:
Force balance: F↑ = PA = F↓ = P0A + mpg ; P0 = 1 bar = 100 kPa
A = (π/4) D2 = (π/4) × 0.1252 = 0.01227 m2
mp = (P-P0)A/g = ( 1500 − 100 ) × 1000 × 0.01227 / 9.80665 = 1752 kg
2.15 A barometer to measure absolute pressure shows a mercury column height of 725
mm. The temperature is such that the density of the mercury is 13550 kg/m3. Find
the ambient pressure.
Solution:
Hg : ∆l = 725 mm = 0.725 m; ρ = 13550 kg/m3
P = ρ g∆l = 13550 × 9.80665 × 0.725 × 10-3 = 96.34 kPa
2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gun-
powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is
the acceleration of the ball if the cylinder (cannon) is pointing horizontally?
Solution:
The cannon ball has 101 kPa on the side facing the atmosphere.
ma = F = P1 × A - P0 × A
a = (P1 - P0 ) × A / m = ( 7000 - 101 ) π [ ( 0.15
2 /4 )/5 ] = 24.38 m/s2
2.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative
to the horizontal direction.
Solution:
ma = F = ( P1 - P0 ) A - mg sin 40
0 
ma = ( 7000 - 101 ) × π × ( 0.152 / 4 ) - 5 × 9.80665 × 0.6428
 = 121.9 - 31.52 = 90.4 N
a = 90.4 / 5 = 18.08 m/s2
2-6
2.18 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg
resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure
of 100 kPa, what should the water pressure be to lift the piston?
Solution:
Force balance: F↑ = F↓ = PA = mpg + P0A
P = P0 + mpg/A = 100 kPa + (100 × 9.80665) / (0.01 × 1000)
 = 100 kPa + 98.07 = 198 kPa
2.19 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what
pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700
kg of a car?
Solution:
F↓ = ma = mg = 740 × 9.80665 = 7256.9 N
Force balance: F↑ = ( P - P0 ) A = F↓ => P = P0 + F↓ / A
A = π D2 (1 / 4) = 0.031416 m2
P = 101 + 7256.9 / (0.031416 × 1000) = 332 kPa
2.20 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local
barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside
the vessel.
Solution:
Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa
P = Pgauge + P0 = 1250 + 96 = 1346 kPa
2.21 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is
97 kPa. If a U-tube with mercury, density 13550 kg/m3, is attached to the tank to
measure the vacuum, what column height difference would it show?
Solution:
∆P = P0 - Ptank = ρg∆l
∆l = ( P0 - Ptank ) / ρg = [(97 - 85 ) × 1000 ] / (13550 × 9.80665)
 = 0.090 m = 90 mm
2-7
2.22 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear spring
and the outside atmosphericpressure of 100 kPa. The spring exerts no force on the
piston when it is at the bottom of the cylinder and for the state shown, the pressure
is 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing the
piston to rise 2 cm. Find the new pressure.
Solution:
A linear spring has a force linear proportional to displacement. F = k x, so
the equilibrium pressure then varies linearly with volume: P = a + bV, with an
intersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V
-> 0) when there is no spring force F = PA = PoA + mpg and the initial state.
These two points determine the straight line shown in the P-V diagram.
Piston area = AP = (π/4) × 0.1
2 = 0.00785 m2
400 
106.2 
2 
1 
0 0.4 
P 
V 
0.557 
2 P 
a = P0 + 
mpg
Ap
 = 100 + 
5 × 9.80665
0.00785
 = 106.2 kPa intersect for zero volume.
V2 = 0.4 + 0.00785 × 20 = 0.557 L
P2 = P1 + 
dP
dV
 ∆V
 = 400 + 
(400-106.2)
0.4 - 0
 (0.557 - 0.4)
 = 515.3 kPa
2.23 A U-tube manometer filled with water, density 1000 kg/m3, shows a height
difference of 25 cm. What is the gauge pressure? If the right branch is tilted to
make an angle of 30° with the horizontal, as shown in Fig. P2.23, what should the
length of the column in the tilted tube be relative to the U-tube?
Solution:
h 
H 
30° 
∆P = F/A = mg/A = Vρg/A = hρg
 = 0.25 × 1000 × 9.807 = 2452.5 Pa
 = 2.45 kPa
 h = H × sin 30°
 ⇒ H = h/sin 30° = 2h = 50 cm
2-8
2.24 The difference in height between the columns of a manometer is 200 mm with a
fluid of density 900 kg/m3. What is the pressure difference? What is the height
difference if the same pressure difference is measured using mercury, density
13600 kg/ m3, as manometer fluid?
Solution:
∆P = ρ1gh1 = 900 × 9.807 × 0.2 = 1765.26 Pa = 1.77 kPa
hhg = ∆P/ (ρhg g) = (ρ1 gh1) / (ρhg g) = 
900
13600
 ×0.2 = 0.0132 m= 13.2 mm
2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercury
manometer. Reservoir A is moved up/down so the two top surfaces are level at h3
as shown in Fig. P2.25. Assuming that you know ρA, ρHg and measure the
heights h1, h2 , and h3, find the density ρB.
Solution:
Balance forces on each side:
P0 + ρAg(h3 - h2) + ρHggh2 = P0 + ρBg(h3 - h1) + ρHggh1
⇒ ρB = ρA




h3 - h2
h3 - h1
 + ρHg




h2 - h1
h3 - h1
2.26 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3,
the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall
with diameter 4m. What is the total force from the bottom of each tank to the water
and what is the pressure at the bottom of each tank?
Solution:
VA = H × πD2 × (1 / 4) = 10 × π × 22
 × ( 1 / 4) = 31.416 m3 
VB = H × πD2 × (1 / 4) = 2.5 × π × 42 × ( 1 / 4) = 31.416 m3
Tanks have the same volume, so same mass of water
F = mg = ρ V g = 1000 × 31.416 × 9.80665 = 308086 N
Tanks have same net force up (holds same m in gravitation field)
Pbot = P0 + ρ H g
Pbot,A = 101 + (1000 × 10 × 9.80665 / 1000) = 199 kPa
Pbot,B = 101 + (1000 × 2.5 × 9.80665 / 1000) = 125.5 kPa
2-9
2.27 The density of mercury changes approximately linearly with temperature as
ρHg = 13595 - 2.5 T kg/ m
3 T in Celsius
so the same pressure difference will result in a manometer reading that is
influenced by temperature. If a pressure difference of 100 kPa is measured in the
summer at 35°C and in the winter at -15°C, what is the difference in column height
between the two measurements?
Solution:
∆P = ρgh ⇒ h = ∆P/ρg ; ρsu = 13507.5 ; ρw = 13632.5
hsu = 100×10
3/(13507.5 × 9.807) = 0.7549 m
hw = 100×10
3/(13632.5 × 9.807) = 0.7480 m
∆h = hsu - hw = 0.0069 m = 6.9 mm
2.28 Liquid water with density ρ is filled on top of a thin piston in a cylinder with cross-
sectional area A and total height H. Air is let in under the piston so it pushes up,
spilling the water over the edge. Deduce the formula for the air pressure as a
function of the piston elevation from the bottom, h.
Solution:
Force balance
H 
h 
P 0 Piston: F↑ = F↓
PA = P0A + mH2O
g
P = P0 + mH2O
g/A
 P = P0 + (H-h)ρg
h, V air 
P 
P 0 
2.29 A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The
setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of
piston motion towards the gas. Assuming standard atmospheric pressure outside
the cylinder, find the gas pressure.
Solution:
P 0 
g 
Force balance: F↑ = F↓ = P0A + mpg = PA
P = P0 + mpg/A = 101.325 + 5 × 25 / (1000 × 0.0015)
 = 184.7 kPa
2-10
2.30 A piece of experimental apparatus is located where g = 9.5 m/s2 and the
temperature is 5°C. An air flow inside the apparatus is determined by measuring
the pressure drop across an orifice with a mercury manometer (see Problem 2.27
for density) showing a height difference of 200 mm. What is the pressure drop in
kPa?
Solution:
∆P = ρgh ; ρHg = 13600
∆P = 13600 × 9.5 × 0.2 = 25840 Pa = 25.84 kPa
2.31 Repeat the previous problem if the flow inside the apparatus is liquid water, ρ ≅
1000 kg/m3, instead of air. Find the pressure difference between the two holes
flush with the bottom of the channel. You cannot neglect the two unequal water
columns.
Solution: Balance forces in the manometer:
P P 1 2 · · 
h 
h 1 
2 
H 
(H - h2) - (H - h1) = ∆hHg = h1 - h2
 P1A + ρH2Oh1gA + ρHg(H - h1)gA
 = P2A + ρH2Oh2gA + ρHg(H - h2)gA
⇒ P1 - P2 = ρH2O(h2 - h1)g + ρHg(h1 - h2)g
 P1 - P2 = ρHg∆hHgg - ρH2O∆hHgg = 13600 × 0.2 × 9.5 - 1000 × 0.2 × 9.5
 = 25840 - 1900 = 23940 Pa = 23.94 kPa
2.32 Two piston/cylinder arrangements, A and B, have their gas chambers connected by
a pipe. Cross-sectional areas are AA = 75 cm
2 and AB = 25 cm2 with the piston
mass in A being mA = 25 kg. Outside pressure is 100 kPa and standard gravitation.
Find the mass mB so that none of the pistons have to rest on the bottom.
Solution: 
A B 
P P 0 0 
Force balance for both pistons: F↑ = F↓
 A: mPAg + P0AA = PAA
 B: mPBg + P0AB = PAB
Same P in A and B gives no flow between them.
 
mPAg
AA 
 + P0 = 
mPBg
AB
 + P0
 => mPB = mPA AA/ AB = 25 × 25/75 = 8.33 kg
2-11
2.33 Two hydraulic piston/cylinders are of same size and setup as in Problem 2.32, but
with neglible piston masses. A single point force of 250 N presses down on piston
A. Find the needed extra force on piston B so that none of the pistons have to
move.
Solution:
No motion in connecting pipe: PA = PB & Forces on pistons balance
AA = 75 cm
2 ; AB = 25 cm
2
PA = P0 + FA / AA = PB = P0 + FB / AB
FB = FA AB / AA = 250 × 25 / 75 = 83.33 N
2.34 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the
ocean and you later climb a hill up to 250 m elevation. Assume the density of water
is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do you
feel at each place?
Solution:
∆P = ρgh
 Pocean= P0 + ∆P = 1025 × 100 + 1000 × 9.81 × 15
 = 2.4965 × 105 Pa = 250 kPa
 Phill = P0 - ∆P = 1025 × 100 - 1.18 × 9.81 × 250
 = 0.99606 × 105 Pa = 99.61 kPa
2.35 In the city water tower, water is pumped up to a level 25 m above ground in a
pressurized tank with air at 125 kPa over the water surface. This is illustrated in
Fig. P2.35. Assuming the water density is 1000 kg/m3 and standard gravity, find
the pressure required to pump more water in at ground level.
Solution:
Pbottom = Ptop + ρgl = 125 + 1000 × 9.807 × 25 × 10
-3
 = 370 kPa
2-12
2.36 Two cylinders are connected by a piston as shown in Fig. P2.36. Cylinder A is
used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and
there is standard gravity. What is the gas pressure in cylinder B?
Solution: 
Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAA
AA = (π/4)0.1
2 = 0.00785 m2; AB = (π/4)0.025
2 = 0.000491 m2
PBAB = PAAA - mpg - P0(AA - AB) = 500× 0.00785 - (25 × 9.807/1000)
 - 100 (0.00785 - 0.000491)= 2.944 kN
PB = 2.944/0.000491 = 5996 kPa = 6.0 MPa
2.37 Two cylinders are filled with liquid water, ρ = 1000 kg/m3, and connected by a line
with a closed valve. A has 100 kg and B has 500 kg of water, their cross-sectional
areas are AA = 0.1 m
2 and AB = 0.25 m
2 and the height h is 1 m. Find the pressure
on each side of the valve. The valve is opened and water flows to an equilibrium.
Find the final pressure at the valve location.
Solution:
VA = vH2O
mA = mA/ρ = 0.1 = AAhA => hA = 1 m
VB = vH2O
mB = mB/ρ = 0.5 = ABhB => hB = 2 m
PVB = P0 + ρg(hB+H) = 101325 + 1000 × 9.81 × 3 = 130 755 Pa
PVA = P0 + ρghA = 101325 + 1000 × 9.81 × 1 = 111 135 Pa
Equilibrium: same height over valve in both
Vtot = VA + VB = h2AA + (h2 - H)AB ⇒ h2 = 
hAAA + (hB+H)AB
AA + AB
 = 2.43 m
PV2 = P0 + ρgh2 = 101.325 + (1000 × 9.81 × 2.43)/1000 = 125.2 kPa
2.38 Using the freezing and boiling point temperatures for water in both Celsius and
Fahrenheit scales, develop a conversion formula between the scales. Find the
conversion formula between Kelvin and Rankine temperature scales.
Solution:
TFreezing = 0 
oC = 32 F; TBoiling = 100 
oC = 212 F
∆T = 100 oC = 180 F ⇒ ToC = (TF - 32)/1.8 or TF = 1.8 ToC + 32
For the absolute K & R scales both are zero at absolute zero.
TR = 1.8 × TK
2-13
English Unit Problems
2.39E A 2500-lbm car moving at 15 mi/h is accelerated at a constant rate of 15 ft/s2 up
to a speed of 50 mi/h. What are the force and total time required?
Solution:
a = 
dV
dt
 = ∆V / ∆ t ⇒ ∆ t = ∆V / a
∆ t = (50 -15) × 1609.34 × 3.28084/(3600 × 15) = 3.42 sec
F = ma = 2500 × 15 / 32.174 lbf= 1165 lbf
2.40E Two pound moles of diatomic oxygen gas are enclosed in a 20-lbm steel
container. A force of 2000 lbf now accelerates this system. What is the
acceleration?
Solution:
mO2
 = nO2
MO2
 = 2 × 32 = 64 lbm
mtot = mO2
 + msteel = 64 + 20 = 84 lbm
a = 
Fgc
mtot
 = (2000 × 32.174) / 84 = 766 ft/s2
2.41E A bucket of concrete of total mass 400 lbm is raised by a crane with an
acceleration of 6 ft/s2 relative to the ground at a location where the local
gravitational acceleration is 31 ft/s2. Find the required force.
Solution:
F = ma = Fup - mg
Fup = ma + mg = 400 × ( 6 + 31 ) / 32.174 = 460 lbf
2.42E One pound-mass of diatomic oxygen (O2 molecular weight 32) is contained in a
100-gal tank. Find the specific volume on both a mass and mole basis (v and v ).
Solution:
v = V/m = 15/1 = 15 ft3/lbm
v̄ = V/n = 
V
m/M
 = Mv = 32 ×15 = 480 ft3/lbmol
2-14
2.43E A 30-lbm steel gas tank holds 10 ft3 of liquid gasoline, having a density of 50
lbm/ft3. What force is needed to accelerate this combined system at a rate of 15
ft/s2?
Solution:
m = mtank + mgasoline = 30 + 10 × 50 = 530 lbm
F = 
ma
gC
 = (530 × 15) / 32.174 = 247.1 lbf
2.44E A differential pressure gauge mounted on a vessel shows 185 lbf/in.2 and a local
barometer gives atmospheric pressure as 0.96 atm. Find the absolute pressure
inside the vessel.
Solution:
P = Pgauge + P0 = 185 + 0.96 × 14.696 = 199.1 lbf/in
2
2.45E A U-tube manometer filled with water, density 62.3 lbm/ft3, shows a height
difference of 10 in. What is the gauge pressure? If the right branch is tilted to
make an angle of 30° with the horizontal, as shown in Fig. P2.23, what should the
length of the column in the tilted tube be relative to the U-tube?
Solution: 
h 
H 
30° 
 ∆P = F/A = mg/AgC = hρg/gC
 = [(10/12) × 62.3 × 32.174] / 32.174 ×144
 = Pgauge = 0.36 lbf/in
2
 h = H × sin 30°
 ⇒ H = h/sin 30° = 2h = 20 in = 0.833 ft
2.46E A piston/cylinder with cross-sectional area of 0.1 ft2 has a piston mass of 200 lbm
resting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressure
of 1 atm, what should the water pressure be to lift the piston?
Solution:
P = P0 + mpg/Agc = 14.696 + (200×32.174) / (0.1 × 144 × 32.174)
 = 14.696 + 13.88 = 28.58 lbf/in2
2-15
2.47E The density of mercury changes approximately linearly with temperature as
 ρHg = 851.5 - 0.086 T lbm/ft
3 T in degrees Fahrenheit
so the same pressure difference will result in a manometer reading that is
influenced by temperature. If a pressure difference of 14.7 lbf/in.2 is measured in
the summer at 95 F and in the winter at 5 F, what is the difference in column
height between the two measurements?
Solution:
∆P = ρgh/gc ⇒ h = ∆Pgc/ρg
ρsu = 843.33 lbm/ft
3; ρw = 851.07 lbm/ft
3
hsu = 
14.7 × 144 × 32.174
843.33 × 32.174
 = 2.51 ft = 30.12 in
hw = 
14.7 × 144 × 32.174
851.07 × 32.174
= 2.487 ft = 29.84 in
∆h = hsu - hw = 0.023 ft = 0.28 in
2.48E A piston, mp = 10 lbm, is fitted in a cylinder, A = 2.5 in.
2, that contains a gas. The
setup is in a centrifuge that creates an acceleration of 75 ft/s2. Assuming standard
atmospheric pressure outside the cylinder, find the gas pressure.
Solution:
P 0 
g 
F↓ = F↑ = P0A + mpg = PA
P = P0 + mpg/Agc = 14.696 + 
10 × 75
2.5 × 32.174
 = 14.696 + 9.324 = 24.02 lbf/in2
2.49E At the beach, atmospheric pressure is 1025 mbar. You dive 30 ft down in the
ocean and you later climb a hill up to 300 ft elevation. Assume the density of
water is about 62.3 lbm/ft3 and the density of air is 0.0735 lbm/ft3. What pressure
do you feel at each place?
Solution:
∆P = ρgh; P0 = (1.025/1.01325) ×14.696 = 14.866 lbf/in
2
Pocean = P0 + ∆P = 14.866 + 
62.3 × 30 × g
gc × 144
 = 27.84 lbf/in2
Phill = P0 -∆P = 14.866 - 
0.0735 × 300 × g
gc × 144
 = 14.71 lbf/in2
3-1
CHAPTER 3
The SI set of problems are revised from the 4th edition as:
New Old New Old New Old
1 new 21 new 41 33
2 new 22 13 mod 42 34
3 new 23 16 mod 43 35
4 new 24 17 44 36 mod
5 new 25 new 45 37 mod
6 new 26 18 mod 46 38 mod
7 7 mod 27 19 d.mod 47 39
8 3 28 20 e.mod 48 40
9 2 29 21 a.b.mod 49 41
10 4 30 22 b.mod 50 42 mod
11 5 31 23 51 43
12 new 32 24 52 44
13 6 33 26 53 45
14 8 mod 34 27 mod 54 46
15 10 35 14 55 47
16 11 36 28 56 48
17 12 37 29 mod 57 49
18 new 38 30 mod 58 50
19 15 mod 39 31 59 51
20 new 40 32 mod 60 52
The english unit problem set is revised from the 4th edition as:
New Old New Old New Old
61 new 69 61 77 69
62 53 70 62 mod 78 70
63 55 71 63 79 71
64 56 72 64 80 72
65 new 73 65 81 new
66 new 74 66 82 74
67 59 mod 75 67 83 75 mod
68 60 76 68
mod indicates a modification from the previous problem that changes the solution
but otherwise is the same type problem.
3-2
3.1 Water at 27°C can exist in different phases dependent upon the pressure. Give the
approximate pressure range in kPa for water being in each one of the three phases
vapor, liquid or solid.
Solution:
The phases can be seen in Fig. 3.6, a sketch
of which is shown to the right.
 T =27 °C = 300 Κ
From Fig. 3.6:
 PVL ≈ 4 × 10
−3 MPa = 4 kPa,
 PLS = 10
3 MPa
ln P
T
V
L
S
CR.P.
P < 4 kPa VAPOR P > 1000 MPa SOLID(ICE)
0.004 MPa < P < 1000 MPa LIQUID
3.2 Find the lowest temperature at which it is possible to have water in the liquid
phase. At what pressure must the liquid exist?
Solution:
 There is no liquid at lower temperatures than on
the fusion line, see Fig. 3.6, saturated ice III to
liquid phase boundary is at
 T ≈ 263K ≈ - 10°C and P ≈ 2100 MPa
ln P
T
V
L
S
CR.P.
lowest T liquid
3.3 If density of ice is 920 kg/m3, find the pressure at the bottom of a 1000 m thick
ice cap on the north pole. What is the melting temperature at that pressure?
Solution: ρICE = 920 kg/m3
∆P = ρgH = 920 × 9.80665 × 1000 = 9022118 Pa
P = Po + ∆P = 101.325 + 9022 = 9123 kPa
See figure 3.6 liquid solid interphase => TLS = -1°C
3-3
3.4 A substance is at 2 MPa, 17°C in a rigid tank. Using only the critical properties
can the phase of the mass be determined if the substance is nitrogen, water or
propane ?
Solution: Find state relative to critical point properties which are:
a) Nitrogen N2 : 3.39 MPa 126.2 K
b) Water H2O : 22.12 MPa 647.3 K
c) Propane C3H8 : 4.25 MPa 369.8 K
 
State is at 17 °C = 290 K and 2 MPa < Pc
for all cases:N2 : T >> Tc Superheated vapor P < Pc
H2O : T << Tc ; P << Pc
 you cannot say.
C3H8 : T < Tc ; P < Pc you cannot say
ln P
T
Vapor
Liquid Cr.P.
a
c
b
3.5 A cylinder fitted with a frictionless piston contains butane at 25°C, 500 kPa. Can
the butane reasonably be assumed to behave as an ideal gas at this state ?
Solution Butane 25°C, 500 kPa, Table A.2: Tc = 425 K; Pc = 3.8 MPa 
Tr = ( 25 + 273 ) / 425 = 0.701; Pr = 0.5/3.8 = 0.13
Look at generalized chart in Figure D.1
Actual Pr > Pr, sat => liquid!! not a gas
3.6 A 1-m3 tank is filled with a gas at room temperature 20°C and pressure 100 kPa.
How much mass is there if the gas is a) air, b) neon or c) propane ?
Solution: Table A.2 T= 20 °C = 293.15 K ; P = 100 kPa << Pc for all
Air : T >> TC,N2; TC,O2 = 154.6 K so ideal gas; R= 0.287
Neon : T >> Tc = 44.4 K so ideal gas; R = 0.41195
Propane: T < Tc = 370 K, but P << Pc = 4.25 MPa so gas R = 0.18855
a) m = PV/RT = 100 ×1 / 0.287 × 293.15 = 1.189 kg
b) m = 100 × 1/ 0.41195 × 293.15 = 0.828 kg
c) m = 100 × 1 / 0.18855 × 293.15 = 1.809 kg
3-4
3.7 A cylinder has a thick piston initially held by a pin as shown in Fig. P3.7. The
cylinder contains carbon dioxide at 200 kPa and ambient temperature of 290 K.
The metal piston has a density of 8000 kg/m3 and the atmospheric pressure is 101
kPa. The pin is now removed, allowing the piston to move and after a while the
gas returns to ambient temperature. Is the piston against the stops?
Solution:
Force balance on piston determines equlibrium float pressure.
Piston mp = Ap × l × ρ ρpiston = 8000 kg/m
3
Pext on CO2
 = P0 + 
mpg
Ap
 = 101 + 
Ap × 0.1 × 9.807 × 8000
Ap × 1000
 = 108.8 kPa
Pin released, as P1 > Pfloat piston moves up, T2 = To & if piston at stops,
then V2 = V1 × 150 / 100
⇒ P2 = P1 × V1 / V2 = 200 × 
100
150
 = 133 kPa > Pext
⇒ piston is at stops, and P2 = 133 kPa
3.8 A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then
filled with carbon dioxide gas at 25°C. To what pressure should it be charged if there
should be 1.2 kg of carbon dioxide?
Solution:
Assume CO2 is an ideal gas table A.5: P = mRT/V
Vcyl = A × L = 
π
4
(0.2)2 × 1 = 0.031416 m3
⇒ P = 1.2 × 0.18892 (273.15 + 25)/0.031416 = 2152 kPa
3-5
3.9 A 1-m3 rigid tank with air at 1 MPa, 400 K is connected to an air line as shown in
Fig. P3.9. The valve is opened and air flows into the tank until the pressure reaches
5 MPa, at which point the valve is closed and the temperature inside is 450K.
a. What is the mass of air in the tank before and after the process?
b. The tank eventually cools to room temperature, 300 K. What is the pressure
inside the tank then?
Solution:
P, T known at both states and assume the air behaves as an ideal gas.
m
air1
 = 
P
1
V
RT
1
 = 
1000 × 1
0.287 × 400
 = 8.711 kg
m
air2
 = 
P
2
V
RT
2
 = 
5000 × 1
0.287 × 450
 = 38.715 kg
Process 2 → 3 is constant V, constant mass cooling to T
3
P
3
 = P
2
 × (T
3
/T
2
) = 5000 × (300/450) = 3.33 MPa
3.10 A hollow metal sphere of 150-mm inside diameter is weighed on a precision
beam balance when evacuated and again after being filled to 875 kPa with an
unknown gas. The difference in mass is 0.0025 kg, and the temperature is 25°C.
What is the gas, assuming it is a pure substance listed in Table A.5 ?
Solution:
Assume an ideal gas with total volume: V = 
π
6
(0.15)3 = 0.001767 m3
M = 
 _ 
mRT
PV
 = 
0.0025 × 8.3145 × 298.2
875 × 0.001767
 = 4.009 ≈ M
He
=> Helium Gas
3-6
3.11 A piston/cylinder arrangement, shown in Fig. P3.11, contains air at 250 kPa,
300°C. The 50-kg piston has a diameter of 0.1 m and initially pushes against the
stops. The atmosphere is at 100 kPa and 20°C. The cylinder now cools as heat is
transferred to the ambient.
a. At what temperature does the piston begin to move down?
b. How far has the piston dropped when the temperature reaches ambient?
Solution:
Piston Ap = 
π
4
 × 0.12 = 0.00785 m2
Balance forces when piston floats:
Pfloat = Po + 
mpg
Ap
 = 100 + 
50 × 9.807
0.00785 × 1000
 = 162.5 kPa = P2 = P3
To find temperature at 2 assume ideal gas:
2
1
P
V
P2
Vstop
3
 T2 = T1 × 
P2
P1
 = 573.15 × 
162.5
250
 = 372.5 K
b) Process 2 -> 3 is constant pressure as piston floats to T3 = To = 293.15 K
V2 = V1 = Ap × H = 0.00785 × 0.25 = 0.00196 m
3 = 1.96 L
Ideal gas and P2 = P3 => V3 = V2 × 
T3
T2
 = 1.96 × 
293.15
372.5
 = 1.54 L
∆H = (V2 -V3)/A = (1.96-1.54) × 0.001/0.00785 = 0.053 m = 5.3 cm
3.12 Air in a tank is at 1 MPa and room temperature of 20°C. It is used to fill an
initially empty balloon to a pressure of 200 kPa, at which point the diameter is 2
m and the temperature is 20°C. Assume the pressure in the balloon is linearly
proportional to its diameter and that the air in the tank also remains at 20°C
throughout the process. Find the mass of air in the balloon and the minimum
required volume of the tank.
Solution: Assume air is an ideal gas.
Balloon final state: V2 = (4/3) π r
3 = (4/3) π 23 = 33.51 m3
 m2bal = P2 V2 / RT2 = 200× 33.51 / 0.287 × 293.15 = 79.66 kg
Tank must have P2 ≥ 200 kPa => m2 tank ≥ P2 VTANK /RT2
Initial mass must be enough: m1 = m2bal + m2 tank = P1V1 / R T1
P1VTANK / R T1 = m2bal + P2VTANK / RT2 =>
VTANK = RTm2bal / (P1 – P2) = 0.287 × 293.15 × 79.66/ (1000 – 200 )
 = 8.377 m3
3-7
3.13 A vacuum pump is used to evacuate a chamber where some specimens are dried at
50°C. The pump rate of volume displacement is 0.5 m3/s with an inlet pressure of
0.1 kPa and temperature 50°C. How much water vapor has been removed over a 30-
min period?
Solution:
Use ideal gas P << lowest P in steam tables. R is from table A.5
m = m
.
 ∆t with mass flow rate as: m
.
= V
.
/v = PV
.
/RT (ideal gas)
⇒ m = PV
.
∆t/RT = 
0.1 × 0.5 × 30×60
(0.46152 × 323.15)
 = 0.603 kg
3.14 An initially deflated and flat balloon is connected by a valve to a 12 m3 storage
tank containing helium gas at 2 MPa and ambient temperature, 20°C. The valve is
opened and the balloon is inflated at constant pressure, Po = 100 kPa, equal to
ambient pressure, until it becomes spherical at D1 = 1 m. If the balloon is largerthan this, the balloon material is stretched giving a pressure inside as
P = P0 + C 





1 − 
D1
D
 
D1
D
The balloon is inflated to a final diameter of 4 m, at which point the pressure
inside is 400 kPa. The temperature remains constant at 20°C. What is the
maximum pressure inside the balloon at any time during this inflation process?
What is the pressure inside the helium storage tank at this time?
Solution:
At the end of the process we have D = 4 m so we can get the constant C as
P = 400 = P0 + C ( 1 – 
1
4
 ) 
1
4
 = 100 + C × 3/16 => C = 1600
The pressure is: P = 100 + 1600 ( 1 – X –1) X –1; X = D / D1
Differentiate to find max: 
dP
dD
 = C ( - X –2 + 2 X –3 ) / D1 = 0
 => - X –2 + 2 X –3 = 0 => X = 2
at max P => D = 2D1 = 2 m; V = 
π
6
 D3 = 4.18 m3
Pmax = 100 + 1600 ( 1 - 
1
2
 ) 
1
2
 = 500 kPa
Helium is ideal gas A.5: m = PV / RT = 
500 × 4.189
2.0771 × 293.15
 = 3.44 kg
mTANK, 1 = PV/RT = 2000 × 12/(2.0771 × 293.15) = 39.416 kg
mTANK, 2 = 39.416 – 3.44 = 35.976 kg
PT2 = mTANK, 2 RT/V = ( mTANK, 1 / mTANK, 2 ) × P1 = 1825.5 kPa
3-8
3.15 The helium balloon described in Problem 3.14 is released into the atmosphere and
rises to an elevation of 5000 m, with a local ambient pressure of Po = 50 kPa and
temperature of −20°C. What is then the diameter of the balloon?
Solution:
Balloon of Problem 3.14, where now after filling D = 4 m, we have :
m1 = P1V1/RT1 = 400 (π/6) 43 /2.077 × 293.15 = 22.015 kg
P1 = 400 = 100 + C(1 - 0.25)0.25 => C = 1600
For final state we have : P0 = 50 kPa, T2 = T0 = -20°C = 253.15 K
State 2: T2 and on process line for balloon, i.e. the P-V relation:
P = 50 + 1600 ( D* -1 - D* -2 ), D* = D/D1 ; V =(π/6) D
3
P2V2 = m R T2 = 22.015 × 2.077 × 253.15 = 11575
or PD+3 = 11575 × 6/ π = 22107 substitute P into the P-V relation
22107 D* -3 = 50 + 1600 ( D* -1 - D* -2 ) Divide by 1600
13.8169 D* -3 - 0.03125 - D* -1 + D* -2 = 0 Multiply by D*3
13.8169 - 0.03125 D* 3 - D* 2 + D* 1 = 0 Qubic equation.
By trial and error D* = 3.98 so D = D*D1 = 3.98 m
3.16 A cylinder is fitted with a 10-cm-diameter piston that is restrained by a linear spring
(force proportional to distance) as shown in Fig. P3.16. The spring force constant is
80 kN/m and the piston initially rests on the stops, with a cylinder volume of 1 L.
The valve to the air line is opened and the piston begins to rise when the cylinder
pressure is 150 kPa. When the valve is closed, the cylinder volume is 1.5 L and the
temperature is 80°C. What mass of air is inside the cylinder?
Solution:
Fs = ks∆x = ks ∆V/Ap ; V1 = 1 L = 0.001 m
3, Ap = 
π
4
 0.12 = 0.007854 m2
State 2: V3 = 1.5 L = 0.0015 m
3; T3 = 80°C = 353.15 K
The pressure varies linearly with volume seen from a force balance as:
PAp = P0 Ap + mp g + ks(V - V0)/Ap
 Between the states 1 and 2 only volume varies so:
 P3 = P2 + 
ks(V3-V2)
Ap
2 = 150 + 
80×103(0.0015 - 0.001)
0.0078542 × 1000
 = 798.5 kPa
 m = 
P3V3
RT3
 = 
798.5 × 0.0015
0.287 × 353.15
 = 0.012 kg
P 
v 
2 
3 
1 
3-9
3.17 Air in a tire is initially at −10°C, 190 kPa. After driving awhile, the temperature goes
up to 10°C. Find the new pressure. You must make one assumption on your own.
Solution:
Assume constant volume and that air is an ideal gas
P2 = P1 × T2/T1 = 190 × 283.15/263.15 = 204.4 kPa
3.18 A substance is at 2 MPa, 17°C in a 0.25-m3 rigid tank. Estimate the mass from the
compressibility factor if the substance is a) air, b) butane or c) propane.
Solution: Figure D.1 for compressibility Z and table A.2 for critical properties.
Nitrogen Pr = 2/3.39 = 0.59; Tr = 290/126.2 = 2.3; Z ≈ 0.98
m = PV/ZRT = 2000 × 0.25/(0.98 × 0.2968 × 290) = 5.928 kg
Butane Pr = 2/3.80 = 0.526; Tr = 290/425.2 = 0.682; Z ≈ 0.085
m = PV/ZRT = 2000 × 0.25/(0.085 × 0.14304 × 290) = 141.8 kg
Propane Pr = 2/4.25 = 0.47; Tr = 290/369.8 = 0.784; Z ≈ 0.08
m = PV/ZRT = 2000 × 0.25/(0.08 × 0.18855 × 290) = 114.3 kg
ln Pr
Z
T = 2.0r
a
bc
T = 0.7r
T = 0.7r
0.1 1
3.19 Argon is kept in a rigid 5 m3 tank at −30°C, 3 MPa. Determine the mass using the
compressibility factor. What is the error (%) if the ideal gas model is used?
Solution: No Argon table so we use generalized chart Fig. D.1
Tr = 243.15/150.8 = 1.612, Pr = 3000/4870 = 0.616 => Z ≅ 0.96
m = 
PV
ZRT
 = 
3000 × 5
0.96 × 0.2081 × 243.2
= 308.75 kg
Ideal gas Z = 1
m = PV/RT = 296.4 kg 4% error
3-10
3.20 A bottle with a volume of 0.1 m3 contains butane with a quality of 75% and a
temperature of 300 K. Estimate the total butane mass in the bottle using the
generalized compressibility chart.
Solution:
m = V/v so find v given T1 and x as : v = vf + x vfg
Tr = 300/425.2 = 0.705 => Fig. D.1 Zf ≈ 0.02; Zg ≈ 0.9
P = Psat = Prsat × Pc = 0.1× 3.80 ×1000 = 380 kPa
vf = ZfRT/P = 0.02 × 0.14304 × 300/380 = 0.00226 m3/kg
vg = ZgRT/P = 0.9 × 0.14304 × 300/380 = 0.1016 m3/kg
v = 0.00226 + 0.75 × (0.1016 – 0.00226) = 0.076765 m3/kg
m = 0.1/0.076765 = 1.303 kg
3.21 A mass of 2 kg of acetylene is in a 0.045 m3 rigid container at a pressure of 4.3
MPa. Use the generalized charts to estimate the temperature. (This becomes trial
and error).
Solution:
Table A.2, A.5: Pr = 4.3/6.14 = 0.70; Tc = 308.3; R= 0.3193
v = V/m = 0.045/2 = 0.0225 m3/kg
State given by (P, v) v = ZRT/P
Since Z is a function of the state Fig. D.1 and thus T, we have trial and error.
Try sat. vapor at Pr = 0.7 => Fig. D.1: Zg = 0.59; Tr = 0.94
vg = 0.59 × 0.3193 × 0.94 × 308.3/4300 = 0.0127 too small
Tr = 1 => Z = 0.7 => v = 0.7 × 0.3193 × 1 × 308.3/4300 = 0.016
Tr = 1.2 => Z = 0.86 => v = 0.86 × 0.3193 × 1.2 × 308.3/4300 = 0.0236
Interpolate to get: Tr ≈ 1.17 T ≈ 361 K
3.22 Is it reasonable to assume that at the given states the substance behaves as an ideal
gas?
Solution:
a) Oxygen, O2 at 30°C, 3 MPa Ideal Gas ( T » Tc = 155 K from A.2)
b) Methane, CH4 at 30°C, 3 MPa Ideal Gas ( T » Tc = 190 K from A.2)
c) Water, H2O at 30°C, 3 MPa NO compressed liquid P > Psat (B.1.1)
3-11
d) R-134a at 30°C, 3 MPa NO compressed liquid P > Psat (B.5.1)
e) R-134a at 30°C, 100 kPa Ideal Gas P is low < Psat (B.5.1)
ln P
T
Vapor
Liq.
Cr.P.
a, b
c, d
e
3.23 Determine whether water at each of the following states is a compressed liquid, a
superheated vapor, or a mixture of saturated liquid and vapor.
Solution: All states start in table B.1.1 (if T given) or B.1.2 (if P given)
a. 10 MPa, 0.003 m3/kg
vf = 0.001452; vg = 0.01803 m
3/kg, so mixture of liquid and vapor.
b. 1 MPa, 190°C : T > Tsat = 179.91oC so it is superheated vapor
c. 200°C, 0.1 m3/kg: v < vg = 0.12736 m3/kg, so it is two-phase
d. 10 kPa, 10°C : P > Pg = 1.2276 kPa so compressed liquid
e. 130°C, 200 kPa: P < Pg = 270.1 kPa so superheated vapor
f. 70°C, 1 m3/kg
vf = 0.001023; vg = 5.042 m
3/kg, so mixture of liquid and vapor
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
Ta
d
e
c
b e
d
a c
b
P = const.
f f
3-12
3.24 Determine whether refrigerant R-22 in each of the following states is a
compressed liquid, a superheated vapor, or a mixture of saturated liquid and
vapor.
Solution:
All cases are seen in Table B.4.1
a. 50°C, 0.05 m3/kg superheated vapor, v > vg =0.01167 at 50°C
b. 1.0 MPa, 20°C compressed liquid, P > Pg = 909.9 kPa at 20°C
c. 0.1 MPa, 0.1 m3/kg mixture liq. & vapor, vf < v < vg at 0.1 MPa
d. 50°C, 0.3 m3/kg superheated vapor, v > vg = 0.01167 at 50°C
e −20°C, 200 kPa superheated vapor, P < Pg = 244.8 kPa at -20°C
f. 2 MPa, 0.012 m3/kg superheated vapor, v > vg = 0.01132 at 2 MPa
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
a
d
e
c
b
e
dacb
P = const.
f
f
3.25 Verify the accuracy of the ideal gas model when it is used to calculate specific
volume for saturated water vapor as shown in Fig. 3.9. Do the calculation for 10
kPa and 1 MPa.
Solution:
Look at the two states assuming ideal gas and then the steam tables.
Ideal gas:
v = RT/P => v1 = 0.46152 × (45.81 + 273.15)/10 = 14.72 m3/kg
v2 = 0.46152 × (179.91 + 273.15)/1000 = 0.209 m3/kg
Real gas:
Table B.1.2: v1 = 14.647 m3/kg so error = 0.3 %
v2 = 0.19444 m3/kg so error = 7.49 %
3-13
3.26 Determine the quality (if saturated) or temperature (if superheated) of the
following substances at the given two states:
Solution:
a) Water, H2O, use Table B.1.1 or B.1.2
1) 120°C, 1 m3/kg => v > vg superheated vapor, T = 120 °C
2) 10 MPa, 0.01 m3/kg => two-phase v < vg
x = ( 0.01 – 0.001452 ) / 0.01657 = 0.516
b) Nitrogen, N2, table B.6
1) 1 MPa, 0.03 m3/kg => superheated vapor since v > vg
Interpolate between sat. vapor and superheated vapor B.6.2:
T ≅ 103.73 + (0.03-0.02416)×(120-103.73)/(0.03117-0.02416) = 117 K
2) 100 K, 0.03 m3/kg => sat. liquid + vapor as two-phase v < vg
v = 0.03 = 0.001452 + x × 0.029764 ⇒ x = 0.959
c) Ammonia, NH3, table B.2
1) 400 kPa, 0.327 m3/kg => v > vg = 0.3094 m
3/kg at 400 kPa
Table B.2.2 superheated vapor T ≅ 10 °C
2) 1 MPa, 0.1 m3/kg => v < vg 2-phase roughly at 25 °C
x = ( 0.1 – 0.001658 ) / 0.012647 = 0.7776
d) R-22, table B.4
1) 130 kPa, 0.1 m3/kg => sat. liquid + vapor as v < vg
 vf ≅ 0.000716 m
3/kg, vg ≅ 0.1684 m3/kg
v = 0.1 = 0.000716 + x × 0.16768 ⇒ x = 0.592
2) 150 kPa, 0.17 m3/kg => v > vg superheated vapor, T ≅ 0°C
3.27 Calculate the following specific volumes
Solution:
a. R-134a: 50°C, 80% quality in Table B.5.1
v = 0.000908 + x × 0.014217 = 0.01228 m3/kg
b. Water 4 MPa, 90% quality in Table B.1.2
v = 0.001252(1-x) + x× 0.04978 = 0.04493 m3/kg
c. Methane 140 K, 60% quality in Table B.7.1
v = 0.00265 + x × 0.09574 = 0.06009 m3/kg
d. Ammonia 60°C, 25% quality in Table B.2.1
v = 0.001834 + x × 0.04697 = 0.01358 m3/kg
3-14
3.28 Give the phase and the specific volume.
Solution:
a. H2O T = 275°C P = 5 MPa Table B.1.1 or B.1.2
Psat = 5.94 MPa => superheated vapor v = 0.04141 m
3/kg
b. H2O T = −2°C P = 100 kPa Table B.1.5
Psat = 0.518 kPa =>compressed solid v ≅ vi = 0.0010904 m
3/kg
c. CO2 T = 267°C P = 0.5 MPa Table A.5
sup. vap. assume ideal gas v = 
RT
P
 = 
0.18892 × 540
500
 = 0.204 m3/kg
d. Air T = 20°C P = 200 kPa Table A.5
sup. vap. assume ideal gas v = 
RT
P
 = 
0.287 × 293
200
 = 0.420 m3/kg
e. NH3 T = 170°C P = 600 kPa Table B.2.2
T > Tc => sup. vap. v = (0.34699 + 0.36389)/2 = 0.3554m
3/kg
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
a
c, d, e
b
a
c, d, e
b
P = const.
3.29 Give the phase and the specific volume.
Solution:
a. R-22 T = −25°C P = 100 kPa => Table B.4.1 Psat = 201 kPa
sup. vap. B.4.2 v ≅ (0.22675 + 0.23706)/2 = 0.2319 m3/kg
b. R-22 T = −25°C P = 300 kPa => Table B.4.1 Psat = 201 kPa
compr. liq. as P > Psat v ≅ vf = 0.000733 m
3/kg
c. R-12 T = 5°C P = 300 kPa => Table B.3.1 Psat = 362.6 kPa
sup. vap. B.3.2 v ≅ (0.0569 + 0.05715)/2 = 0.05703 m3/kg
d. Ar T = 200°C P = 200 kPa Table A.5
ideal gas v = 
RT
P
 = 
0.20813 × 473
200
 = 0.4922 m3/kg
3-15
e. NH3 T = 20°C P = 100 kPa => Table B.2.1 Psat = 847.5 kPa
sup. vap. B.2.2 v = 1.4153 m3/kg
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
a, c, e
b
a, c, eb
P = const.
d
d
3.30 Find the phase, quality x if applicable and the missing property P or T.
Solution:
a. H2O T = 120°C v = 0.5 m
3/kg < vg Table B.1.1
sat. liq. + vap. P = 198.5 kPa, x = (0.5 - 0.00106)/0.8908 = 0.56
b. H2O P = 100 kPa v = 1.8 m
3/kg Table B.1.2 v > vg
sup. vap., interpolate in Table B.1.3
 T = 
1.8 − 1.694
1.93636 − 1.694
 (150 – 99.62) + 99.62 = 121.65 °C
c. H2O T = 263 K v = 200 m
3/kg Table B.1.5
sat. solid + vap., P = 0.26 kPa, x = (200-0.001)/466.756 = 0.4285
d. Ne P = 750 kPa v = 0.2 m3/kg; Table A.5
ideal gas, T = 
Pv
R
 = 
750 × 0.2
0.41195
 = 364.1 K
e. NH3 T = 20°C v = 0.1 m
3/kg Table B.2.1
sat. liq. + vap. , P = 857.5 kPa, x = (0.1-0.00164)/0.14758 = 0.666
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
b
a, e b
P = const.
d
d
a, e
c c
3-16
3.31 Give the phase and the missing properties of P, T, v and x.
Solution:
a. R-22 T = 10°C v = 0.01 m3/kg Table B.4.1
 sat. liq. + vap. P = 680.7 kPa, x = (0.01-0.0008)/0.03391 = 0.2713
b. H2O T = 350°C v = 0.2 m3/kg Table B.1.1 v > vg
sup. vap. P ≅ 1.40 MPa, x = undefined
c. CO2 T = 800 K P = 200 kPa Table A.5
ideal gas v = 
RT
P
 = 
0.18892 × 800
200
 = 0.756 m3/kg
d. N2 T = 200 K P = 100 kPa Table B.6.2 T > Tc
sup. vap. v = 0.592 m3/kg
e. CH4 T = 190 K x = 0.75 Table B.7.1 P = 4520 kPa
sat. liq + vap. v = 0.00497 + x × 0.003 = 0.00722 m3/kg
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
b
a, e b
P = const.
c, d
a, e
c, d
3.32 Give the phase and the missing properties of P, T, v and x. These may be a little
more difficult if the appendix tables are used instead of the software.
Solution:
a) R-22 at T = 10°C, v = 0.036 m3/kg: Table B.4.1 v > vg at 10°C
=> sup. vap. Table B.4.2 interpolate between sat. and sup. both at 10°C
P = 680.7 + (0.036-0.03471)(600-680.7)/(0.04018-0.03471) = 661.7 kPa
b) H2O v = 0.2 m3/kg , x = 0.5: Table B.1.1
sat. liq. + vap. v = (1-x) vf + x vg => vf + vg = 0.4 m3/kg
since vf is so small we find it approximately where vg = 0.4 m3/kg.
vf + vg = 0.39387 at 150°C, vf + vg = 0.4474 at 145°C.
An iterpolation gives T ≅ 149.4°C, P ≅ 468.2 kPa
3-17
c) H2O T = 60°C, v = 0.001016 m3/kg: Table B.1.1 v < vf = 0.001017
 => compr. liq. see Table B.1.4
v = 0.001015 at 5 MPa so P ≅ 0.5(5000 + 19.9) = 2.51 MPa
d) NH3 T = 30°C, P = 60 kPa : Table B.2.1 P < Psat
 => sup. vapor interpolate in Table B.2.2
 v = 2.94578 + (60-50)(1.95906-2.94578)/(75-50) = 2.551 m3/kg
v is not linearly proportional to P (more like 1/P) so the computer table
gives a more accurate value of 2.45
e) R-134a v = 0.005m3/kg , x = 0.5: sat. liq. + vap. Table B.5.1
v = (1-x) vf + x vg => vf + vg = 0.01 m3/kg
vf + vg = 0.010946 at 65°C, vf + vg = 0.009665 at 70°C.
An iterpolation gives: T ≅ 68.7°C, P = 2.06 MPa
States shown are
placed relative to the
two-phase region, not
to each other.
P C.P.
v
T
C.P.
v
T
a
b, e
a
P = const.
db, ec
c
d
3.33 What is the percent error in specific volume if the ideal gas model is used to
represent the behavior of superheated ammonia at 40°C, 500 kPa? What if the
generalized compressibility chart, Fig. D.1, is used instead?
Solution:
NH3 T = 40°C = 313.15 K, Tc = 405.5 K, Pc = 11.35 MPa from Table A.1
Table B.2.2: v = 0.2923 m3/kg
Ideal gas: v = 
RT
P
 = 
0.48819 × 313
500
 = 0.3056 m3/kg ⇒ 4.5% error
Figure D.1: Tr = 313.15/405.5 = 0.772, Pr = 0.5/11.35 = 0.044 ⇒ Z = 0.97
v = ZRT/P = 0.2964 m3/kg ⇒ 1.4% error
3-18
3.34 What is the percent error in pressure if the ideal gas model is used to represent the
behavior of superheated vapor R-22 at 50°C, 0.03082 m3/kg? What if the
generalized compressibility chart, Fig. D.1, is used instead (iterations needed)?
Solution: Real gas behavior: P = 900 kPa from Table B.4.2
Ideal gas constant: R = R
_
/M = 8.31451/86.47 = 0.096155
P = RT/v = 0.096155 × (273.15 + 50) / 0.03082
 = 1008 kPa which is 12% too high
Generalized chart Fig D.1 and critical properties from A.2:
Tr = 323.2/363.3 = 0.875; Pc = 4970 kPa
Assume P = 900 kPa => Pr = 0.181 => Z ≅ 0.905
v = ZRT/P = 0.905 × 0.096155 × 323.15 / 900 = 0.03125 too high
Assume P = 950 kPa => Pr = 0.191 => Z ≅ 0.9
v = ZRT/P = 0.9 × 0.096155 × 323.15 / 950 = 0.029473 too low
P ≅ 900 + ( 950 − 900 ) × 
0.03082 − 0.029437
0.03125 − 0.029437
 = 938 kPa 4.2 % high
3.35 Determine the mass of methane gas stored in a 2 m3 tank at −30°C, 3 MPa. Estimate
the percent error in the mass determination if the ideal gas model is used.
Solution:
The methane Table B.7.2 linear interpolation between 225 and 250 K.
⇒ v ≅ 0.03333 + 
243.15-225
250-225
×(0.03896-0.03333) = 0.03742 m3/kg
m = V/v = 2/0.03742 = 53.45 kg
Ideal gas assumption
v = RT/P = 0.51835 × 243.15/3000 = 0.042
m = V/v = 2/0.042 = 47.62 kg
Error: 5.83 kg 10.9% too small
3.36 A water storage tank contains liquid and vapor in equilibrium at 110°C. The distance
from the bottom of the tank to the liquid level is 8 m. What is the absolute pressure
at the bottom of the tank?
Solution:
Saturated conditions from Table B.1.1: Psat = 143.3 kPa
vf = 0.001052 m
3/kg ; ∆P = 
gh
vf
 = 
9.807 × 8
0.001052
 = 74578 Pa = 74.578 kPa
Pbottom = Ptop + ∆P = 143.3 + 74.578 = 217.88 kPa
3-19
3.37 A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C.
The vessel is now heated. If a safety pressure valve is installed, at what pressure
should the valve be set to have a maximum temperature of 200°C?
Solution:
Process: v = V/m = constant
 v1 = 1/2 = 0.5 m
3/kg 2-phase
200°C, 0.5 m3/kg seen in Table B.1.3 to be
between 400 and 500 kPa so interpolate
 P ≅ 400 + 
0.5-0.53422
 0.42492-0.53422
 × (500-400)
 = 431.3 kPa
C.P.T
v
100 C
500 kPa
400 kPa
3.38 A 500-L tank stores 100 kg of nitrogen gas at 150 K. To design the tank the
pressure must be estimated and three different methods are suggested. Which is
the most accurate, and how different in percent are the other two?
a. Nitrogen tables, Table B.6
b. Ideal gas
c. Generalized compressibilitychart, Fig. D.1
Solution:
State 1: 150 K, v = V/m = 0.5/100 = 0.005 m3/kg
a) Table B.6, interpolate between 3 & 6 MPa with both at 150 K:
3 MPa : v = 0.01194 6 MPa : v = 0.0042485
P= 3 + (0.005-0.01194)×(6-3)/(0.0042485-0.01194) = 5.707 MPa
b) Ideal gas table A.5: P = 
RT
v
 = 
0.2968 × 150
0.005
 = 8.904 MPa
c) Table A.2 Tc = 126.2 K, Pc = 3.39 MPa so Tr = 150/126.2 = 1.189
 Z is a function of P so it becomes trial and error. Start with P = 5.7 MPa
Pr ≅ 1.68 ⇒ Z = 0.60 ⇒ P = 
ZRT
v
 = 5342 kPa
⇒ Pr = 1.58 ⇒ Z = 0.62 ⇒ P = 5520 kPa OK
ANSWER: a) is the most accurate with others off by b) 60% c) 1%
3-20
3.39 A 400-m3 storage tank is being constructed to hold LNG, liquified natural gas,
which may be assumed to be essentially pure methane. If the tank is to contain
90% liquid and 10% vapor, by volume, at 100 kPa, what mass of LNG (kg) will
the tank hold? What is the quality in the tank?
Solution:
CH4 at P = 100 kPa from Table B.7.1 by interpolation.
mliq = 
Vliq
vf
 = 
0.9 × 400
0.00236
 = 152542 kg; mvap = 
Vvap
vg
 = 
0.1 × 400
0.5726
 = 69.9 kg
mtot = 152 612 kg, x = mvap / mtot = 4.58×10
-4
(If you use computer table, vf ≅ 0.002366, vg ≅ 0.5567)
3.40 A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up
by 5°C per hour due to a failure in the refrigeration system. How long time will it
take before the methane becomes single phase and what is the pressure then?
Solution: Use Table B.7.1
Assume rigid tank v = const = v1 = 0.002439 + 0.25×0.30367 = 0.078366
All single phase when v = vg => T ≅ 145 K
∆t = ∆Τ/5°C ≅ (145 – 120 ) / 5 = 5 hours P = Psat= 824 kPa
3.41 Saturated liquid water at 60°C is put under pressure to decrease the volume by 1%
keeping the temperature constant. To what pressure should it be compressed?
Solution: H2O T = 60°C , x = 0.0; Table B.1.1
v = 0.99 × vf (60°C) = 0.99×0.001017 = 0.0010068 m
3/kg
Between 20 & 30 MPa in Table B.1.4, P ≅ 23.8 MPa
3.42 Saturated water vapor at 60°C has its pressure decreased to increase the volume
by 10% keeping the temperature constant. To what pressure should it be
expanded?
Solution:
From initial state: v = 1.10 × vg = 1.1 × 7.6707 = 8.4378 m
3/kg
Interpolate at 60°C between saturated (P = 19.94 kPa) and superheated vapor
P = 10 kPa in Tables B.1.1 and B.1.3
P ≅ 19.941 + (8.4378 − 7.6707)(10-19.941)/(15.3345-7.6707) = 18.9 kPa
Comment: T,v ⇒ P = 18 kPa (software) v is not linear in P, more like 1/P,
so the linear interpolation in P is not very accurate.
3-21
3.43 A boiler feed pump delivers 0.05 m3/s of water at 240°C, 20 MPa. What is the mass
flowrate (kg/s)? What would be the percent error if the properties of saturated liquid
at 240°C were used in the calculation? What if the properties of saturated liquid at
20 MPa were used?
Solution:
At 240°C, 20 MPa: v = 0.001205 m3/kg (from B.1.4)
m
.
 = V
.
/v = 0.05/0.001205 = 41.5 kg/s
vf (240°C) = 0.001229 ⇒ m
.
 = 40.68 kg/s error 2%
vf (20 MPa) = 0.002036 ⇒ m
.
 = 24.56 kg/s error 41%
3.44 A glass jar is filled with saturated water at 500 kPa, quality 25%, and a tight lid is
put on. Now it is cooled to −10°C. What is the mass fraction of solid at this
temperature?
Solution:
Constant volume and mass ⇒ v1 = v2 From Table B.1.2 and B.1.5:
v1 = 0.001093 + 0.25 × 0.3738 = 0.094543 = v2 = 0.0010891 + x2 × 446.756
⇒ x2 = 0.0002 mass fraction vapor
xsolid =1-x2 = 0.9998 or 99.98 %
3.45 A cylinder/piston arrangement contains water at 105°C, 85% quality with a
volume of 1 L. The system is heated, causing the piston to rise and encounter a
linear spring as shown in Fig. P3.45. At this point the volume is 1.5 L, piston
diameter is 150 mm, and the spring constant is 100 N/mm. The heating continues,
so the piston compresses the spring. What is the cylinder temperature when the
pressure reaches 200 kPa?
Solution:
P1 = 120.8 kPa, v1 = vf + x vfg = 0.001047 + 0.85*1.41831 = 1.20661
m = V1/ v1 = 
0.001
1.20661
 = 8.288×10-4 kg
v2 = v1 (V2 / V1) = 1.20661× 1.5 = 1.8099
& P = P1 = 120.8 kPa ( T2 = 203.5°C )
P3 = P2 + (ks/Ap
2) m(v3-v2) linear spring
P
v
1 2 3
200
1 1.5 liters
Ap = (π/4) × 0.15
2 = 0.01767 m2 ; ks = 100 kN/m (matches P in kPa)
200 = 120.8 + (100/0.01767 2 ) × 8.288×10-4(v3-1.8099)
200 = 120.8 + 265.446 (v3 – 1.8099)=> v3 = 2.1083 m
3/kg
T3 ≅ 600 + 100 × (2.1083 – 2.01297)/(2.2443-2.01297) ≅ 641°C
3-22
3.46 Saturated (liquid + vapor) ammonia at 60°C is contained in a rigid steel tank. It is
used in an experiment, where it should pass through the critical point when the
system is heated. What should the initial mass fraction of liquid be?
Solution:
Process: Constant mass and volume, v = C
From table B.2.1:
v1 = v2 = 0.004255 = 0.001834 + x1 × 0.04697
 => x1 = 0.01515
 liquid = 1 - x1 = 0.948
T
v
60 C
Crit. point
1
3.47 For a certain experiment, R-22 vapor is contained in a sealed glass tube at 20°C. It is
desired to know the pressure at this condition, but there is no means of measuring it,
since the tube is sealed. However, if the tube is cooled to −20°C small droplets of
liquid are observed on the glass walls. What is the initial pressure?
Solution: R-22 fixed volume (V) & mass (m) at 20°C cool to -20°C ~ sat. vapor
1 
1 
 
2 
20 C
o 
20 C
o - 
P
T
T
T
v
 v = const = vg at -20°C = 0.092843 m
3/kg
 State 1: 20°C, 0.092843 m3/kg
interpolate between 250 and 300 kPa in
Table B.4.2
 => P = 291 kPa
3-23
3.48 A steel tank contains 6 kg of propane (liquid + vapor) at 20°C with a volume of
0.015 m3. The tank is now slowly heated. Will the liquid level inside eventually rise
to the top or drop to the bottom of the tank? What if the initial mass is 1 kg instead
of 6 kg?
Solution: Constant volume and mass v2 = v1 = V/m = 0.0025 m
3/kg
v 
T 
20°C
C.P.
V c 
vc = 0.203/44.094 = 0.004604 > v1
 eventually reaches sat. liq.
 ⇒ level rises to top
If m = 1 kg ⇒ v1 = 0.015 > vc
 then it will reach sat. vap.
 ⇒ level falls
3.49 A cylinder containing ammonia is fitted with a piston restrained by an external force
that is proportional to cylinder volume squared. Initial conditions are 10°C, 90%
quality and a volume of 5 L. A valve on the cylinder is opened and additional
ammonia flows into the cylinder until the mass inside has doubled. If at this point the
pressure is 1.2 MPa, what is the final temperature?
Solution:
State 1 Table B.2.1: v1 = 0.0016 + 0.9(0.205525 - 0.0016) = 0.18513 m
3/kg
P1 = 615 kPa; V1 = 5 L = 0.005 m
3
m1 = V/v = 0.005/0.18513 = 0.027 kg
State 2: P2 = 1.2 MPa, Flow in so: m2 = 2 m1 = 0.054 kg
Process: Piston Fext = KV
2 = PA => P = CV2 => P2 = P1 (V2/V1)
2
From the process equation we then get:
V2 = V1 (P2/P1)
1/2
 = 0.005 (
1200
615
)
1/2
 = 0.006984 m3
v2 = V/m = 
0.006984
0.054
 = 0.12934 m3/kg
At P2, v2: T2 = 70.9°C
3-24
3.50 A container with liquid nitrogen at 100 K has a cross sectional area of 0.5 m2.
Due to heat transfer, some of the liquid evaporates and in one hour the liquid level
drops 30 mm. The vapor leaving the container passes through a valve and a heater
and exits at 500 kPa, 260 K. Calculate the volume rate of flow of nitrogen gas
exiting the heater.
Solution:
Properties from table B.6.1 for volume change, exit flow from table B.6.2:
 ∆V = A × ∆h = 0.5 × 0.03 = 0.015 m3
 ∆mliq = -∆V/vf = -0.015/0.001452 = -10.3306 kg
 ∆mvap = ∆V/vg = 0.015/0.0312 = 0.4808 kg
 mout = 10.3306 - 0.4808 = 9.85 kg
 vexit = 0.15385 m
3/kg
 V
.
 = m
.
vexit = (9.85 / 1h)× 0.15385 = 1.5015 m
3/h = 0.02526 m3/min
3.51 A pressure cooker (closed tank) contains water at 100°C with the liquid volume
being 1/10 of the vapor volume. It is heated until the pressure reaches 2.0 MPa.
Find the final temperature. Has the final state more or less vapor than the initial
state?
Solution:
Vf = mf vf = Vg/10 = mgvg/10 ; vf = 0.001044, vg = 1.6729x1 = 
mg
mg + mf
 = 
10 mfvf / vg
mf + 10 mfvf / vg
 = 
10 vf
10 vf + vg
 = 
0.01044
0.01044 + 1.6729
 = 0.0062
v1 = 0.001044 + 0.0062×1.67185 = 0.01141 = v2 < vg(2MPa) so two-phase
0.01141 = 0.001177 + x2 × 0.09845 => x2 = 0.104 More vapor
T2 = Tsat(2MPa) = 212.4°C
3-25
3.52 Ammonia in a piston/cylinder arrangement is at 700 kPa, 80°C. It is now cooled at
constant pressure to saturated vapor (state 2) at which point the piston is locked with
a pin. The cooling continues to −10°C (state 3). Show the processes 1 to 2 and 2 to 3
on both a P–v and T–v diagram.
Solution:
700 
290 
P 
v 
2 
3 
1 
v 
T 
2 
3 
1 80
14
-10 
3.53 A piston/cylinder arrangement is loaded with a linear spring and the outside
atmosphere. It contains water at 5 MPa, 400°C with the volume being 0.1 m3. If the
piston is at the bottom, the spring exerts a force such that Plift = 200 kPa. The
system now cools until the pressure reaches 1200 kPa. Find the mass of water, the
final state (T2, v2) and plot the P–v diagram for the process.
P 
v 
5000
1200
200 
1 
2 
a 
? 0.05781 0 
1: Table B.1.3 ⇒ v1= 0.05781
 m = V/v1 = 0.1/0.05781 = 1.73 kg
Straight line: P = Pa + Cv
v2 = v1 
P2 - Pa
P1 - Pa
 = 0.01204 m3/kg
v2 < vg(1200 kPa) so two-phase T2 = 188°C
 ⇒ x2 = (v2 - 0.001139)/0.1622 = 0.0672
3-26
3.54 Water in a piston/cylinder is at 90°C, 100 kPa, and the piston loading is such that
pressure is proportional to volume, P = CV. Heat is now added until the temperature
reaches 200°C. Find the final pressure and also the quality if in the two-phase region.
Solution:
 Final state: 200°C , on process line P = CV
1
2
v
P
State 1: Table B.1.1: v1 = 0.001036 m
3/kg
 P2 = P1v2/v1 from process equation
Check state 2 in Table B.1.1
 vg(T2) = 0.12736; Pg(T2) = 1.5538 MPa
If v2 = vg(T2) ⇒ P2 = 12.3 MPa > Pg not OK
If sat. P2 = Pg(T2) = 1553.8 kPa ⇒ v2 = 0.0161 m
3kg < vg sat. OK,
P2 = 1553.8 kPa, x2 = (0.0161 - 0.001156) / 0.1262 = 0.118
3.55 A spring-loaded piston/cylinder contains water at 500°C, 3 MPa. The setup is
such that pressure is proportional to volume, P = CV. It is now cooled until the
water becomes saturated vapor. Sketch the P-v diagram and find the final
pressure.
Solution:
2 
1 
P
v
P = Cv ⇒ C = P1/v1 = 3000/0.11619 = 25820
State 2: x2 = 1 & P2 = Cv2 (on process line)
Trial & error on T2sat or P2sat:
at 2 MPa vg = 0.09963 ⇒ C = 20074
 2.5 MPa vg = 0.07998 ⇒ C = 31258
 2.25 MPa vg = 0.08875 ⇒ C = 25352
Interpolate to get right C ⇒ P2 = 2270 kPa
3.56 Refrigerant-12 in a piston/cylinder arrangement is initially at 50°C, x = 1. It is then
expanded in a process so that P = Cv−1 to a pressure of 100 kPa. Find the final
temperature and specific volume.
Solution:
State 1: 50°C, x=1 ⇒ P1 = 1219.3 kPa, v1 = 0.01417 m
3/kg
Process: Pv = C = P1v1; => P2 = C/v2= P1v1/v2
State 2: 100 kPa and v2 = v1P1/P2 = 0.1728 m
3/kg
T2 ≅ -13.2°C from Table B.3.2
3-27
3.57 A sealed rigid vessel of 2 m3 contains a saturated mixture of liquid and vapor R-
134a at 10°C. If it is heated to 50°C, the liquid phase disappears. Find the
pressure at 50°C and the initial mass of the liquid.
Solution:
 Process: constant volume and constant mass.
P
v
2
1
State 2 is saturated vapor, from table B.5.1
 P2 = Psat(50°C) = 1.318 MPa
State 1: same specific volume as state 2
v1 = v2 = 0.015124 m
3/kg
v1 = 0.000794 + x1 × 0.048658
 ⇒ x1 = 0.2945
 m = V/v1 = 2/0.015124 = 132.24 kg; mliq = (1 - x1)m = 93.295 kg
3.58 Two tanks are connected as shown in Fig. P3.58, both containing water. Tank A is at
200 kPa, v = 0.5 m3/kg, VA = 1 m
3 and tank B contains 3.5 kg at 0.5 MPa, 400°C.
The valve is now opened and the two come to a uniform state. Find the final specific
volume.
Solution:
Control volume: both tanks. Constant total volume and mass process.
mA = VA/vA = 1/0.5 = 2 kg
vB = 0.6173 ⇒ VB = mBvB = 3.5 × 0.6173 = 2.1606 m
3
Final state: mtot = mA + mB = 5.5 kg
Vtot = VA + VB = 3.1606 m
3
v2 = Vtot/mtot = 0.5746 m
3/kg
3-28
3.59 A tank contains 2 kg of nitrogen at 100 K with a quality of 50%. Through a volume
flowmeter and valve, 0.5 kg is now removed while the temperature remains constant.
Find the final state inside the tank and the volume of nitrogen removed if the
valve/meter is located at
a. The top of the tank
b. The bottom of the tank
Solution
m2 = m1 - 0.5 = 1.5 kg
v1 = 0.001452 + x1 × 0.029764 = 0.016334
Vtank = m1v1 = 0.0327 m
3
v2 = Vtank/m2 = 0.0218 < vg(T)
x2 = 
0.0218-0.001452
0.031216-0.001452
 = 0.6836
Top: flow out is sat. vap. vg = 0.031216
Vout = moutvg = 0.0156 m
3
Bottom: flow out is sat. liq. vf = 0.001452
Vout = moutvf = 0.000726 m
3
3.60 Consider two tanks, A and B, connected by a valve, as shown in Fig. P3.60. Each
has a volume of 200 L and tank A has R-12 at 25°C, 10% liquid and 90% vapor
by volume, while tank B is evacuated. The valve is now opened and saturated
vapor flows from A to B until the pressure in B has reached that in A, at which
point the valve is closed. This process occurs slowly such that all temperatures
stay at 25°C throughout the process. How much has the quality changed in tank A
during the process?
mA1 = 
Vliq1
vf 25°C
 + 
Vvap1
vg 25°C
 = 
0.1 × 0.2
0.000763
 + 
0.9 × 0.2
0.026854
 = 26.212 + 6.703 = 32.915 kg
xA1 = 
6.703
32.915
 = 0.2036 ; mB2 = 
VB
vg 25°C
 = 
0.2
0.26854
 = 7.448 kg
⇒ mA2 = 32.915 - 7.448 = 25.467 kg
vA2 = 
0.2
25.467
 = 0.007853 = 0.000763 + xA2 × 0.026091
xA2 = 0.2718 ∆x = 6.82%
3-29
English Unit Problems
3.61E A substance is at 300 lbf/in.2, 65 F in a rigid tank. Using only the critical
properties can the phase of the mass be determined if the substance is nitrogen,
water or propane?
Solution: Find state relative to the critical point properties, table C.1
 Nitrogen 492 lbf/in.2 227.2 R
 Water 3208 lbf/in.2 1165.1 R
 Propane 616 lbf/in.2 665.6 R
 P < Pc for all and T = 65 F =65 + 459.67 = 525 R
 N2 T >> Tc Yes gas and P < Pc
 H2O T << Tc P << Pc so you cannot say
 C3H8 T < Tc P < Pc you cannot say
3.62E A cylindrical gas tank 3 ft long, inside diameter of 8 in., is evacuated and then
filled with carbon dioxide gas at 77 F. To what pressure should it be charged if
there should be 2.6 lbm of carbon dioxide?
Solution:
Assume CO2 is an ideal gas table C.4: P = mRT/V
Vcyl = A × L = 
π
4
 (8)2 × 3 × 12 = 1809.6 in3
P = 
2.6 × 35.1 × (77 + 459.67) × 12
1809.6
 = 324.8 lbf/in2
3.63E A vacuum pump is used to evacuate a chamber where some specimens are dried at
120 F. The pump rate of volume displacement is 900 ft3/min with an inlet pressure
of 1 mm Hg and temperature 120 F. How much water vapor has been removed over
a 30-min period?
Solution:
Use ideal gas as P << lowest P in steam tables. R is from table C.4
P = 1 mmHg = 0.01934 lbf/in2
m
.
= 
PV
.
RT
 ⇒ m = 
PV
.
∆t
RT
 = 
0.01934 × 900 × 30 × 144
85.76 × (120 + 459.67)
 = 1.513 lbm
3-30
3.64E A cylinder is fitted with a 4-in.-diameter piston that is restrained by a linear spring
(force proportional to distance) as shown in Fig. P3.16. The spring force constant
is 400 lbf/in. and the piston initially rests on the stops, with a cylinder volume of
60 in.3. The valve to the air line is opened and the piston begins to rise when the
cylinder pressure is 22 lbf/in.2. When the valve is closed, the cylinder volume is
90 in.3 and the temperature is 180 F. What mass of air is inside the cylinder?
Solution: V1 = V2 = 60 in
3; Ap = 
π
4
 × 42 = 12.566 in2
P2 = 22 lbf/in
2 ; V3 = 90 in
3 , T3 = 180°F = 639.7 R
Linear spring: P3 = P2 + 
ks(V3-V2)
Ap
2
 = 22 + 
400
12.5662
 (90-60) = 98 lbf/in2
P 
v 
2 
3 
1 
m = 
P3V3
RT3
 = 
98 × 90
12 × 53.34 × 639.7
 = 0.02154 lbm
3.65E A substance is at 70 F, 300 lbf/in.2 in a10 ft3 tank. Estimate the mass from the
compressibility chart if the substance is a) air, b) butane or c) propane.
Solution:
Use Fig. D.1 for compressibility Z and table C.1 for critical properties
m =PV/ZRT = 300 ×144 ×10 / 530 Z R = 815.09 / Z R = 815.09 / Z R
Air use nitrogen 492 lbf/in.2; 227.2 R Pr = 0.61; Tr = 2.33; Z = 0.98
m =PV/ZRT = 815.09 / Z R = 815.09/(0.98× 55.15) = 15.08 lbm
Butane 551 lbf/in.2; 765.4 R Pr = 0.544; Tr = 0.692; Z = 0.09
m =PV/ZRT = 815.09 / Z R = 815.09 /(0.09× 26.58) = 340.7 lbm
Propane 616 lbf/in.2; 665.6 R Pr = 0.487; Tr = 0.796; Z = 0.08
m =PV/ZRT = 815.09 / Z R = 815.09 / (0.08× 35.04) = 290.8 lbm
3.66E Determine the mass of an ethane gas stored in a 25 ft3 tank at 250 F, 440 lbf/in.2
using the compressibility chart. Estimate the error (%) if the ideal gas model is used.
Solution
Table C.1: Tr = ( 250 + 460 ) / 549.7 = 1.29 and Pr = 440/708 = 0.621
Figure D.1 ⇒ Z = 0.9
m = PV/ZRT = 440 × 144 × 25 / (51.38 × 710 × 0.9) = 48.25 lbm
Ideal gas Z = 1 ⇒ m = 43.21 lbm 10% error
3-31
3.67E Argon is kept in a rigid 100 ft3 tank at −30 F, 450 lbf/in.2. Determine the mass
using the compressibility factor. What is the error (%) if the ideal gas model is
used?
Solution: Use the generalized chart in Fig. D.1 and critical values from C.1.
Tr = ( 460 – 30 ) / 271.4 = 1.58, Pr = 450/706 = 0.64 ⇒ Z = 0.95
m = PV/ZRT = 450 × 144 × 100 / (0.95 × 38.68 × 430) = 410 lbm
Ideal gas Z = 1 ⇒ m = PV/RT = 390 lbm 5% error
3.68E Determine whether water at each of the following states is a compressed liquid, a
superheated vapor, or a mixture of saturated liquid and vapor.
Solution: All cases can be seen from Table C.8.1
a. 1800 lbf/in.2, 0.03 ft3/lbm
vg = 0.2183, vf = 0.02472 ft
3/lbm, so liq + vap. mixture
b. 150 lbf/in.2, 320 F: compr. liquid P > Psat(T) = 89.6 lbf/in
2
c. 380 F, 3 ft3/lbm: sup. vapor v > vg(T) = 2.339 ft
3/lbm
d. 2 lbf/in.2, 50 F: compr. liquid P > Psat(T) = 0.178
e. 270 F, 30 lbf/in.2: sup. vapor P < Psat(T) = 41.85 lbf/in
2
f. 160 F, 10 ft3/lbm
vg = 77.22, vf = 0.0164 ft
3/lbm, so liq. + vap. mixture
3.69E Give the phase and the specific volume.
Solution:
a. H2O T = 520F P = 700 lbf/in.
2 Table C.8.1
Psat = 811.5 => sup. vapor v = 0.6832 ft3/lbm
b. H2O T = 30 F P = 15 lbf/in.
2 Table C.8.4
 Psat = 0.0886 => compr. solid v = vi = 0.01747 ft
3/lbm
c. CO2 T = 510 F P = 75 lbf/in.
2 Table C.4
sup. vap. ideal gas v = RT/P = 
35.1 × (510 + 459.7)
75 × 144
 = 3.152 ft3/lbm
d. Air T = 68 F P = 2 atm Table C.4
sup. vap. ideal gas v = RT/P = 
53.34 × (68 + 459.7)
2 × 14.6 × 144
 = 6.6504 ft3/lbm
e. NH3 T = 290 F P = 90 lbf/in.
2 Table C.9.2
sup. vap. v = 4.0965 ft3/lbm
3-32
3.70E Give the phase and the specific volume.
Solution:
a. R-22 T = -10 F, P = 30 lbf/in.2 Table C.10.1 P < Psat = 31.2 psia
 ⇒ sup.vap. v ≅ 1.7439 + 
-10+11.71
 11.71
 (1.7997 – 1.7439) = 1.752 ft3/lbm
b. R-22 T = -10 F, P = 40 lbf/in.2 Table C.10.1 Psat = 31.2 psia
 P > Psat ⇒ compresssed Liquid v ≅ vf = 0.01178 ft
3/lbm
c. H2O T = 280 F, P = 35 lbf/in.
2 Table C.8.1 P < Psat = 49.2 psia
 ⇒ sup.vap v ≅ 21.734 + ( 10.711 – 21.734) ×(15/20) = 1.0669 ft3/lbm
d. Ar T = 300 F, P = 30 lbf/in.2 Table C.4
 Ideal gas: v = RT/P = 38.68 (300 + 459.7) / (30 × 144) = 6.802 ft3/lbm
e. NH3 T = 60 F, P = 15 lbf/in.
2 Table C.9.1 Psat = 107.6 psia
 P < Psat ⇒ sup.vap v ≅ 21.564 ft
3/lbm
3.71E Give the phase and the missing properties of P, T, v and x. These may be a little
more difficult if the appendix tables are used instead of the software.
Solution:
a. R-22 at T = 50 F, v = 0.6 ft3/lbm: Table C.10.1 v > vg
sup. vap. C.10.2 interpolate between sat. and sup. vap at 50F.
P ≅ 98.73 + (0.6 - 0.5561)(80 -98.73)/(0.708 - 0.5561) = 93.3 lbf/in2
b. H2O v = 2 ft
3/lbm, x = 0.5: Table C.8.1
since vf is so small we find it approximately where vg = 4 ft3/lbm.
vf + vg = 4.3293 at 330 F, vf + vg = 3.80997 at 340 F.
linear interpolation T ≅ 336 F, P ≅ 113 lbf/in2
c. H2O T = 150 F, v = 0.01632 ft
3/lbm: Table C.8.1, v < vf
compr. liquid P ≅ 500 lbf/in2
d. NH3 T = 80 F, P = 13 lbf/in.
2 Table C.9.1 P < Psat
sup. vap. interpolate between 10 and 15 psia: v = 26.97 ft3/lbm
v is not linear in P (more like 1/P) so computer table is more accurate.
e. R-134a v = 0.08 ft3/lbm, x = 0.5: Table C.11.1
since vf is so small we find it approximately where vg = 0.16 ft3/lbm.
vf + vg = 0.1729 at 150 F, vf + vg = 0.1505 at 160 F.
linear interpolation T ≅ 156 F, P ≅ 300 lbf/in2
3-33
3.72E What is the percent error in specific volume if the ideal gas model is used to
represent the behavior of superheated ammonia at 100 F, 80 lbf/in.2? What if the
generalized compressibility chart, Fig. D.1, is used instead?
Solution:
Ammonia Table C.9.2: v = 4.186 ft3/lbm
Ideal gas v = 
RT
P
 = 
90.72 × 559.7
80 × 144
 = 4.4076 ft3/lbm 5.3% error
Generalized compressibility chart and Table C.1
Tr = 559.7/729.9 = 0.767, Pr = 80/1646 = 0.0486 => Z ≅ 0.96
v = ZRT/P = 0.96 × 4.4076 = 4.231 ft3/lbm 1.0% error
3.73E A water storage tank contains liquid and vapor in equilibrium at 220 F. The
distance from the bottom of the tank to the liquid level is 25 ft. What is the
absolute pressure at the bottom of the tank?
Solution:
Table C.8.1: vf = 0.01677 ft
3/lbm
∆P = 
g l
gcvf
 = 
32.174 × 25
32.174 × 0.01677 × 144
 = 10.35 lbf/in2
3.74E A sealed rigid vessel has volume of 35 ft3 and contains 2 lbm of water at 200 F.
The vessel is now heated. If a safety pressure valve is installed, at what pressure
should the valve be set to have a maximum temperature of 400 F?
Solution:
Process: v = V/m = constant = v1 = 17.5 ft
3/lbm
Table C.8.2: 400 F, 17.5 ft3/lbm ⇒ between 20 & 40 lbf/in2
P ≅ 32.4 lbf/in2 (28.97 by software)
3.75E Saturated liquid water at 200 F is put under pressure to decrease the volume by
1%, keeping the temperature constant. To what pressure should it be compressed?
Solution:
v = 0.99 × vf = 0.99 × 0.016634 = 0.016468 ft
3/lbm
Table C.8.4 P ~ 3200 lbf/in2
3-34
3.76E Saturated water vapor at 200 F has its pressure decreased to increase the volume by
10%, keeping the temperature constant. To what pressure should it be expanded?
Solution:
v = 1.1 × vg = 1.1 × 33.63 = 36.993 ft
3/lbm
Interpolate between sat. at 200 F and sup. vapor in Table C.8.2 at
200 F, 10 lbf/in2 P ≅ 10.54 lbf/in2
3.77E A boiler feed pump delivers 100 ft3/min of water at 400 F, 3000 lbf/in.2. What is
the mass flowrate (lbm/s)? What would be the percent error if the properties of
saturated liquid at 400 F were used in the calculation? What if the properties of
saturated liquid at 3000 lbf/in.2 were used?
Solution: Table C.8.4: v = 0.0183 ft3/lbm
m
.
 = 
V
.
v
 = 
100
60 × 0.018334
 = 91.07 lbm/s
vf (400 F) = 0.01864 ⇒ m
.
 = 89.41 error 1.8%
vf (3000 lbf/in2) = 0.03475 ⇒ m
.
 = 47.96 error 47%
3.78E Saturated (liquid + vapor) ammonia at 140 F is contained in a rigid steel tank. It is
used in an experiment, where it should pass through the critical point when the
system is heated. What should the initial mass fraction of liquid be?
Solution:
P process constant volume & mass. From Table C.9.1:
v1 = vc = 0.031532 ft
3/lbm = 0.01235 + x1 × 1.1398 => x1 = 0.01683
Liquid fraction = 1 - x1 = 0.983
3-35
3.79E A steel tank contains 14 lbm of propane (liquid + vapor) at 70 F with a volume of
0.25 ft3. The tank is now slowly heated. Will the liquid level inside eventually rise
to the top or drop to the bottom of the tank? What if the initial mass is 2 lbm
instead of 14 lbm?
Solution:
P 
v 
Constant volume and mass
v2 = v1 = V/m = 0.25/14 = 0.01786
vc = 3.2/44.097 = 0.07256 ft
3/lbm
v2 < vc so eventually sat. liquid
 ⇒ level rises
If v2 = v1 = 0.25/2 = 0.125 > vc
Now sat. vap. is reached so leveldrops
3.80E A pressure cooker (closed tank) contains water at 200 F with the liquid volume
being 1/10 of the vapor volume. It is heated until the pressure reaches 300 lbf/in.2.
Find the final temperature. Has the final state more or less vapor than the initial
state?
Solution:
Process: Constant volume and mass.
Vf = mf vf = Vg/10 = mgvg/10; Table C.8.1: vf = 0.01663, vg = 33.631
x1 = 
mg
mg + mf
 = 
10 mfvf / vg
mf + 10 mfvf / vg
 = 
10 vf
10 vf + vg
 = 
0.1663
0.1663 + 33.631
 = 0.00492
v2 = v1 = 0.01663 + x1 × 33.615 = 0.1820 ft
3/lbm
P2, v2 ⇒ T2 = Tsat = 417.43 F
0.1820 = 0.01890 + x2 × 1.5286
x2 = 0.107 more vapor than state 1.
3-36
3.81E Two tanks are connected together as shown in Fig. P3.58, both containing water.
Tank A is at 30 lbf/in.2, v = 8 ft3/lbm, V = 40 ft3 and tank B contains 8 lbm at 80
lbf/in. 2, 750 F. The valve is now opened and the two come to a uniform state.
Find the final specific volume.
Solution:
Control volume both tanks. Constant total volume and mass.
mA = VA/vA = 40/8 = 5 lbm
Table C.8.2: vB = (8.561 + 9.322)/2 = 8.9415
 ⇒ VB = mBvB = 8 × 8.9415 = 71.532 ft
3
Final state: mtot = mA + mB = 5 + 8 = 13 lbm
Vtot = VA + VB = 111.532 ft
3
v2 = Vtot/mtot = 111.532/13 = 8.579 ft
3/lbm
3.82E A spring-loaded piston/cylinder contains water at 900 F, 450 lbf/in.2. The setup is
such that pressure is proportional to volume, P = CV. It is now cooled until the
water becomes saturated vapor. Find the final pressure.
Solution:
2 
P 
v 
1 
State 1: v
1
 = 1.7524 ft3/lbm
P = Cv ⇒ C = P
1
/v
1
 = 256.79
State 2: sat. vap. x
2
 = 1
Trial & error on T
2
 or P
2
At 350 lbf/in2: P
g
/v
g
 = 263.8 > C
At 300 lbf/in2: P
g
/v
g
 = 194.275 < C
Interpolation: P
2
 ≅ 345 lbf/in2
3-37
3.83E Refrigerant-22 in a piston/cylinder arrangement is initially at 120 F, x = 1. It is
then expanded in a process so that P = Cv−1 to a pressure of 30 lbf/in.2. Find the
final temperature and specific volume.
Solution:
State 1: P
1
 = 274.6 lbf/in2 v
1
 = 0.1924 ft3/lbm
Process: Pv = C = P
1
v
1
 = P
2
v
2
State 2: P
2
 = 30 lbf/in2 and on process line (equation).
 v
2
 = 
v
1
P
1
P
2
 = 0.1924 × 274.6/30 = 1.761 ft3/lbm
Table C.10.2 between saturated at -11.71 F and 0 F: T
2
 ≅ -8.1 F
4-1
CHAPTER 4
The new problem set relative to the problems in the fourth edition.
New Old New Old New Old
1 new 21 21 41 41
2 new 22 22 42 42
3 4 23 23 43 43 mod
4 5 24 24 mod 44 32
5 1 25 25 45 16
6 2 26 new 46 33
7 7 27 27 47 new HT
8 new 28 26 mod 48 new HT
9 new 29 28 49 new HT
10 new 30 29 50 new HT
11 8 31 30 51 new HT
12 9 32 new 52 new HT
13 new 33 14 53 new HT
14 11 34 34 54 new HT
15 new 35 35 55 new HT
16 13 36 37 56 new HT
17 15 37 38 57 new HT
18 31 38 20 58 44
19 18 39 39 59 45
20 new 40 40 60 46
61 47 mod
English unit problems
62 48 68 57 74 59
63 new 69 new 75 60 mod
64 49 70 54 76 61
65 50 71 55 mod 77 new
66 new 72 56 78 new
67 52 73 new 79 new
4-2
4.1 A piston of mass 2 kg is lowered 0.5 m in the standard gravitational field. Find the
required force and work involved in the process.
Solution:
F = ma = 2 × 9.80665 = 19.61 N
W = ∫ F dx = F ∫ dx = F ∆x = 19.61 × 0.5 = 9.805 J
4.2 An escalator raises a 100 kg bucket of sand 10 m in 1 minute. Determine the total
amount of work done and the instantaneous rate of work during the process.
Solution:
W = ∫ F dx = F ∫ dx = F ∆x = 100 × 9.80665 × 10 = 9807 J
.
W = W / ∆t = 9807 / 60 = 163 W
4.3 A linear spring, F = ks(x − xo), with spring constant ks = 500 N/m, is stretched
until it is 100 mm longer. Find the required force and work input.
Solution:
F = ks(x - x0) = 500 × 0.1 = 50 N
W = ∫ F dx = ⌡⌠ ks(x - x0)d(x - x0) = ks(x - x0)2/2
 = 500 × 0.12/2 = 2.5 J
4.4 A nonlinear spring has the force versus displacement relation of F = kns(x − xo)
n.
If the spring end is moved to x1 from the relaxed state, determine the formula for
the required work.
Solution:
W = ⌡⌠Fdx = ⌡
⌠
 kns(x - x0)
n d(x - x0) = 
kns
n + 1
 (x1 - x0)
n+1
4-3
4.5 A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant
R-134a vapor at 1000 kPa, 140°C. The setup is cooled at constant pressure until
the R-134a reaches a quality of 25%. Calculate the work done in the process.
Solution:
Constant pressure process boundary work. State properties from Table B.5.2
State 1: v = 0.03150 m3/kg ,
State 2: v = 0.000871 + 0.25 × 0.01956 = 0.00576 m3/kg
Interpolated to be at 1000 kPa, numbers at 1017 kPa could have
been used in which case: v = 0.00566
W12 = ∫ P dV = P (V2-V1) = mP (v2-v1)
 = 5 × 1000(0.00576 - 0.03150) = -128.7 kJ
4.6 A piston/cylinder arrangement shown in Fig. P4.6 initially contains air at 150 kPa,
400°C. The setup is allowed to cool to the ambient temperature of 20°C.
a. Is the piston resting on the stops in the final state? What is the final
pressure in the cylinder?
b. What is the specific work done by the air during this process?
Solution:
P1 = 150 kPa, T1 = 400°C = 673.2 K
T2 = T0 = 20°C = 293.2 K
For all states air behave as an ideal gas.
a) If piston at stops at 2, V2 = V1/2
 and pressure less than Plift = P1
V
P
1
2
1a
P
P
2
1
 ⇒ P2 = P1 × 
V1
V2
 × 
T2
T1
 = 150 × 2 × 
293.2
673.2
 = 130.7 kPa < P1
 ⇒ Piston is resting on stops.
b) Work done while piston is moving at const Pext = P1.
 W12 = ∫ Pext dV = P1 (V2 - V1) ; V2 = 
1
2
 V1 = 
1
2
 m RT1/P1
w12 = W12/m = RT1 (
1
2
 - 1 ) = -
1
2
 × 0.287 × 673.2 = -96.6 kJ/kg
4-4
4.7 The refrigerant R-22 is contained in a piston/cylinder as shown in Fig. P4.7, where
the volume is 11 L when the piston hits the stops. The initial state is −30°C, 150
kPa with a volume of 10 L. This system is brought indoors and warms up to 15°C.
a. Is the piston at the stops in the final state?
b. Find the work done by the R-22 during this process.
Solution:
Initially piston floats, V < Vstop so the
piston moves at constant Pext = P1 until
it reaches the stops or 15°C, whichever is
first.
a) From Table B.4.2: v1 = 0.1487,
 m = V/v = 
0.010
0.1487
 = 0.06725 kg
1a
2
1
P
V
P1
Vstop
 Check the temperature at state 1a: P1a = 150 kPa, v = Vstop/m.
 v2 = V/m = 
0.011
0.06725
 = 0.16357 m3/kg => T1a = -9 °C & T2 = 15°C
 Since T2 > T1a then it follows that P2 > P1 and the piston is againts stop.
 b) Work done at const Pext = P1.
 W12 = ∫ Pext dV = Pext(V2 - V1) = 150(0.011 - 0.010) = 0.15 kJ
4.8 Consider a mass going through a polytropic process where pressure is directly
proportional to volume (n = − 1). The process start with P = 0, V = 0 and ends with P
= 600 kPa, V = 0.01 m3.The physical setup could be as in Problem 2.22. Find the
boundary work done by the mass.
Solution:
The setup has a pressure that varies linear with volume going through the
initial and the final state points. The work is the area below the process curve.
0.01
600
P
V
0
0
W
W = ⌡⌠ PdV = AREA
 = 
1
2 (P1 + P2)(V2 - V1)
 = 
1
2 (P2 + 0)( V2 - 0)
 = 
1
2 P2 V2 = 
1
2 × 600 × 0.01 = 3 kJ
4-5
4.9 A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m3.
Stops in the cylinder restricts the enclosed volume to 0.5 m3, similar to the setup
in Problem 4.7. The water is now heated to 200°C. Find the final pressure,
volume and the work done by the water.
Solution:
Initially the piston floats so the equilibrium
lift pressure is 200 kPa
1: 200 kPa, v1= 0.1/50 = 0.002 m3/kg,
2: 200 °C, ON LINE
Check state 1a: vstop = 0.5/50 = 0.01 =>
Table B.1.2: 200 kPa , vf < vstop < vg
1a
2
1
P
V
P1
Vstop
State 1a is two phase at 200 kPa and Tstop ≈ 120.2 °C so as T2 > Tstop the
state is higher up in the P-V diagram with v2 = vstop < vg = 0.127 (at 200°C)
State 2 two phase => P2 = Psat(T2) = 1.554 MPa, V2 = Vstop = 0.5 m3
1W2 = 1Wstop = 200 (0.5 – 0.1) = 80 kJ
4.10 A piston/cylinder contains 1 kg of liquid water at

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